audio read-through Solving Exponential Equations (Part 2)

(This page is Part 2. Click here for Part 1.)

Example 2

Solve:    $\displaystyle\frac{2^{3t-1}}{7\cdot 5^{2+t}} = 1$

$\displaystyle\frac{2^{3t-1}}{7\cdot 5^{2+t}} = 1$
Original Equation
$2^{3t-1} = 7\cdot 5^{2+t}$
Isolate

Multiply both sides by $\,7\cdot 5^{2+t}\,$ to get in the form ‘ExpTerm1 = ExpTerm2’. Note that $\,7\cdot 5^{2+t}\,$ is never equal to zero, so this equation is equivalent to the former.

$\ln (2^{3t-1}) = \ln (7\cdot 5^{2+t})$
Undo

Take natural logs of both sides. Note that both ‘$\, 2^{3t-1}\,$’ and ‘$\,7\cdot 5^{2+t}\,$’ are always positive, so this equation is equivalent to the former.

$(3t-1)(\ln 2) = \ln 7 + (2+t)(\ln 5)$
Undo (continued)

Use properties of logarithms to get the variables out of the exponents. This is now a linear equation in one variable.

$3t\ln 2 - t\ln 5 = \ln 7 + 2\ln 5 + \ln 2$
Solve

Get all the variable terms on the left, and constant terms on the right.

$t(3\ln 2 - \ln 5) = \ln 7 + 2\ln 5 + \ln 2$
Solve (continued)

Factor out the $\,t\,$ on the LHS.

$\displaystyle t = \frac{\ln 7 + 2\ln 5 + \ln 2}{3\ln 2 - \ln 5}$
Solve (continued)

Divide by $\,3\ln 2 - \ln 5\,.$

$\displaystyle \begin{align} t &= \frac{\ln 7 + 2\ln 5 + \ln 2}{3\ln 2 - \ln 5}\cr\cr &\approx 12.46359 \end{align}$



$\displaystyle\frac{2^{3(12.46359)-1}}{7\cdot 5^{2+12.46359}}\overset{\text{?}}{=} 1$

$1.00000 = 1$

Okay!
Check

Approximate the exact answer to at least five decimal places. Substitute in the original equation. Put a question mark over the equal sign, since you're asking a question—are the two sides equal?

Here, the left-hand side evaluates to the number $\,1\,$ when rounded to five decimal places.

Example 3

Solve:    $5^{x}\cdot 3^{x-1} = \frac 12 7^{2-x}$

$5^{x}\cdot 3^{x-1} = \frac 12 7^{2-x}$
Original Equation
$\ln \bigl(5^{x}\cdot 3^{x-1}\bigr) = \ln\bigl(\frac 12 7^{2-x}\bigr)$
Take logs.
$x\ln 5 + (x-1)\ln 3 = \ln(2^{-1}) + (2-x)\ln 7$
Use properties of logs.
$x(\ln 5 + \ln 3 + \ln 7) = -\ln 2 + \ln 3 + 2\ln 7$
Get variable terms on left and constants on right.
$\displaystyle\begin{align} x &= \frac{-\ln 2 + \ln 3 + 2\ln 7}{(\ln 5 + \ln 3 + \ln 7)} \cr\cr &\approx 0.92336 \end{align}$
Get an exact answer first, and then approximate.
$5^{0.92336}\cdot 3^{0.92336-1} \overset{\text{?}}{=} \frac 12 7^{2-0.92336}$

$4.06288 \approx 4.06290$

Okay!
Check in original equation. Approximate the solution to more than five decimal places, as needed and desired.

Example:  A ‘Fake Quadratic’

Solve:    ${\text{e}}^{2x} - {\text{e}}^x - 6 = 0$

This final exponential equation cannot be put in the form ‘ExpTerm1 = ExpTerm2’. However, it can be turned into a quadratic equation by a simple substitution. Consequently, it is often called a ‘fake quadratic’ or a ‘pseudo-quadratic’ equation.

The idea used is important:

${\text{e}}^{2x} - {\text{e}}^x - 6 = 0$
Original Equation

This equation is not solvable by IUSC.

${(\text{e}}^{x})^2 - ({\text{e}}^x) - 6 = 0$

Use a property of exponents to rename the first term.
A familiar quadratic pattern emerges!

$u^2 - u - 6 = 0$

Let $\,u := {\text{e}}^x\,.$ This substitution transforms the equation in $\,x\,$ to a quadratic equation in $\,u\,.$

$(u-3)(u+2) = 0$

$u = 3$   or   $u = -2$

Solve the transformed equation. You can save a couple steps if you're comfortable never explicitly bringing $\,u\,$ into the picture:

$$ \begin{gather} \cssId{s74}{{(\text{e}}^{x})^2 - ({\text{e}}^x) - 6 = 0}\cr\cr \cssId{s75}{({\text{e}}^{x} - 3)({\text{e}}^{x} + 2) = 0}\cr\cr \cssId{s76}{{\text{e}}^x = 3\ \ \text{ or }\ \ {\text{e}}^x = -2} \end{gather} $$
${\text{e}}^x = 3$   or   ${\text{e}}^x = -2$

Transform back: go back to the original variable, $\,x\,.$

$x\ln \text{e} = \ln 3$

$x = \ln 3 \approx 1.09861$

Since $\,{\text{e}}^x\,$ is always strictly positive, it never equals a negative number. There is only one solution.

${\text{e}}^{2\,\cdot\, 1.09861} - {\text{e}}^{1.09861} - 6 \overset{\text{?}}{=} 0$

$-0.00003 \approx 0$

Okay!

Check.

Concept Practice